Question

a) A large hotel in Miami has 900 rooms (all rooms are equivalent). During Christmas, the hotel is usually fully booked. However, as it is possible for a customer to cancel their reservation, the hotel overbooks its rooms. 1000 people were given assurance of a room. Let us assume that each customer cancels their reservation with a probability of 0.1. If the total number of customers who still keep their booking is more than 900, the hotel has to unfortunately send some customers to other accommodation. What is the probability that this happens, as per the Central Limit Theorem

Answer 1

Answer:

14.69% probability that this happens

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

1000 people were given assurance of a room.

This means that $n = 1000$

Let us assume that each customer cancels their reservation with a probability of 0.1.

So 0.9 probability that they still keep their booking, which means that $p = 0.9$

Probability more than 900 still keeps their booking:

$n = 1000, p = 0.9$

So

$\mu = 0.9, s = \sqrt{\frac{0.9*0.1}{1000}} = 0.0095$

901/1000 = 0.91

So this is 1 subtracted by the pvalue of Z when X = 0.91.

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{0.91 - 0.9}{0.0095}$

$Z = 1.05$

$Z = 1.05$ has a pvalue of 0.8531

1 - 0.8531 = 0.1469

14.69% probability that this happens